Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.1 - The Tangent and Velocity Problems - 2.1 Exercises: 1

Answer

a) At t=5, slope of PQ = $\frac{694-250}{5-15}$ = -$\frac{444}{10}$ = -44.4 At t=10, slope of PQ = $\frac{444-250}{10-15}$ = -$\frac{194}{5}$ = -38.8 At t=20, slope of PQ = $\frac{111-250}{20-15}$ = -$\frac{139}{5}$ = -27.8 At t=25, slope of PQ = $\frac{28-250}{25-15}$ = -$\frac{222}{10}$ = -22.2 At t=30, slope of PQ = $\frac{0-250}{30-15}$ = -$\frac{250}{15}$ = -16.667 b) $\frac{-38.8 + (-27.8)}{2}$ = -33.3 c) Slope of tangent line at P = $\frac{-300}{9}$ = -33.3

Work Step by Step

a) The slope of a secant line PQ = $\frac{x_{2}-x_{1}}{y_{2}-y_{1}}$ when P is ($x_{1}$, $y_{1}$) and Q is ($x_{2}$, $y_{2}$). Since P is the point (15,250), you can substitute (15,250) into the formula so that the slope of a secant line PQ = $\frac{x_{2}-250}{y_{2}-15}$. Then, you plug in the x and y values for Q, which is (5, 694) at t=5; (10, 444) at t= 10; (20, 111) at t=20; (25,28) at t=25; and (30,0) at t=30. b) P is in between t=10 and t=20, so you should average the slopes of t=10 and t=20 to get the most accurate slope for the tangent line at P. To average these two slopes add -38.8 to -27.8 and divide by 2. This should give you an estimate of the slope of the tangent line at P. c) Using the graph of the function, draw a tangent line at P. Plug the x and y values of t=10 and t=20 into the slope formula to find an estimate of the slope of the tangent line at P. After plugging in these values, the formula should look like this: $\frac{111-444}{20-10}$. Finally, simplify and you end up with = -$\frac{300}{9}$ = -33.3
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